Question

the spinner is equally likely to land on any of the six sections. what is the probability that the spinner will land on a number greater than 4 or on a shaded section? 6 9 1 2 2 3 5 6

Explanation:

Step1: Find the number of favorable outcomes

Numbers greater than \(4\) are \(5\) and \(6\). Shaded sections are \(1\), \(3\), \(5\). But \(5\) is counted twice.
Using the formula \(n(A\cup B)=n(A)+n(B)-n(A\cap B)\). Let \(A\) be the set of numbers greater than \(4\) (\(n(A) = 2\)), \(B\) be the set of shaded - section numbers (\(n(B)=3\)), and \(n(A\cap B) = 1\) (the number \(5\)).
So \(n(A\cup B)=2 + 3-1=4\).

Step2: Calculate the probability

The probability \(P=\frac{n(A\cup B)}{n(S)}\), where \(n(S) = 6\) (total number of sections).
So \(P=\frac{4}{6}=\frac{2}{3}\).

Answer:

\(\frac{2}{3}\)